BZOJ1103: [POI2007]大都市meg

题解

应该是个dfs序裸题

土路修成公路时， 把节点子树的权值全部 -1.

查询直接前缀和。

树状数组维护

代码

#include<cstring>
#include<iostream>
#include<cstdio>
#include<algorithm>
#define rd read()
using namespace std;
 
const int N = 5e5;
 
int f[N], head[N], tot, size[N], cnt, id[N];
int sum[N], n, m;
 
struct edge {
    int nxt, to;
}e[N << 1];
 
int read() {
    int X = 0, p = 1; char c = getchar();
    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') p = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0';
    return X * p;
}
 
void added(int u, int v) {
    e[++tot].to = v;
    e[tot].nxt = head[u];
    head[u] = tot;
}
 
void add(int u, int v) {
    added(u, v); added(v, u);
}
 
void dfs(int u) {
    size[u] = 1;
    id[u] = ++cnt;
    for(int i = head[u]; i; i = e[i].nxt) {
        int nt = e[i].to;
        if(nt == f[u]) continue;
        f[nt] = u;
        dfs(nt);
        size[u] += size[nt];
    }
}
 
inline int lowbit(int x) {
    return x & (-x);
}
 
inline void modify(int x, int d) {
    for(; x <= n * 2; x += lowbit(x)) sum[x] += d;
}
 
inline int query(int x) {
    int re = 0;
    for(; x; x -= lowbit(x)) re += sum[x];
    return re;
}
 
int main()
{
    n = rd;
    for(int i = 1; i < n; ++i) {
        int x = rd, y = rd;
        add(x, y);
    }
    dfs(1);
    for(int i = 2; i <= n; ++i) {
        modify(id[i], 1);
        modify(id[i] + size[i], -1);
    }
    m = rd;
    for(int i = 1; i <= n + m - 1; ++i) {
        char k[2]; scanf("%s", k); 
        if(k[0] == 'W') {
            int x = rd;
            int ans = query(id[x]);
            printf("%d
", ans);
        }
        else {
            int x = rd, y = rd;
            if(f[y] == x) swap(x, y);
            modify(id[x], -1);
            modify(id[x] + size[x], 1);
        }
    }
}